DOM has always been notoriously difficult to prepare because of the somewhat odd precursors required and the formylation step needed to make the benzaldehyde. After the discussion in the serious chemistry forum about the two new active compounds DOMCl and DOMOM (

(Lego: "Amphetamines/PEAs w/o benzaldehyde or nitroethane", Novel Discourse), as no gassing is required. I suppose the same yields can be expected from the bromomethylation as the chloromethylation. After the isolation of the -most likely nasty/toxic- DOMCl the next step can commence.

As the whole DOM molecule is quite stable there are many ways of dehalogenate this. Many reducing agents will do the job[1]. The most common being lithium aluminium hydride[2]. Another powerful reagent, which reduces primary, secondary, tertiary, allylic, vinylic, aryl and neopentyl halides, is a complex formed from lithium trimethoxyaluminium hydride LiAlH(OMe)3 and CuI[3]. A milder reducing agent is Sodium Borohydride, NaBH4, in a dipolar aprotic solvent such as Me2SO (DMSO), DMF or sulfolane[4], which at room temperature or above reduces primary, secondary, and some tertiary[5] halides in good yield without affecting other functional groups that would be reduced by LiAlH4.

There are also many other ways of dehalogenating alkylhalides, but many of these seem quite exotic compared to the ones i have mentioned. See [6] for further reading.

For bees the route with borohydride in DMSO / DMF is probably the best way to go. The yields are supposed to be high, and the reaction conditions are mild. An obstacle could be the large aromatic ring connected to the same carbon as the halogen, which could cause some sterical hinderance. However, [4] and [5] states that secondary and even some tertiary halogens are reduceable using this method. I would guess that it ought to work even though it might be a bit hindered with this in mind. I don't have the reference on the specific reaction conditions on this one, but perhaps a friendly bee can dig it up somewhere? After a successfull dehalogenation, we all know what wonderfull molecule awaits us . So if it really is a feasible route i would think that the synthesis of DOM has become quite alot easier. But i suppose there is only one way of finding out if it works nicely

(Rhodium: "Styrene Derivatives by Aromatic Chloroalkylation", Novel Discourse)), might well work with the H2SO4/KBr variation too. Edit: Well, after translating the first reference a bit, it appears that the starting point of the article is the fact that when an aromatic is condensed with acetaldehyde in presence of hydrochloric and sulfuric acids, the product is the corresponding alpha-diphenylethane, not the alpha-chloroethylbenzene.

Concerning the dehalogenation, it occurred to me that the Zn/Ni couple reduction patent (

(moo: "Reduction of phenylacetonitriles w/ Zn/Ni couple", Chemistry Discourse)) also discussed dehalogenation, but the halogen there was aromatic. I guess there isn't much difference compared to traditional catalytic dehalogenations with hydrogen, as nickel is a hydrogenation calalyst like platinum and palladium which are used in those reactions, but of course it is obvious that the benzylic carbon can make a great difference along with the lower activity of the nickel catalyst. And maybe something else as the source of hydrogen just to avoid messy workups. The literature could have something to say about this subject I haven't had a reason to concentrate on yet.

(Nicodem: "Azole, that's a nice and ingenious synth", Serious Chemistry): "However, Bandil forgot that the DOMBr just like DOMCl must not be free-based and to choose the reducing agent appropriately to this.". Nicodem instead suggests tin(II)chloride: "Could the reaction they used for DOMCl be used for the synthesis of DOM, maybe even without isolating the DOMCl intermediate, simply by adding SnCl2×2H2O at the end of the chloromethylation (see

Bandil

God dammit! Why is it so hard to read and understand an article on mondays - sorry about that. Should have read it more carefully before blabbering about.

But of course you are right, acidic reduction it is. If the bromomethylation is done in GAA with sulfuric acid and KBr, the resulting mixture should be quite acidic. Will this environment be adequate for reduction with Stannic(II)Chloride hydrate?

In the reference you gave Rhodium, they use dioxane / HCl as the solvent for the reduction. Perhaps GAA / H2SO4 will work aswell? It could be one sweet reaction if so!

Oh well, i suppse theres really only one way to find out

RegardsBandil

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Rhodium

I think there is a point in using conc. HCl rather than H2SO4 (as SnCl2 is used). The dioxane is probably there for the sake of solubilizing the organics while still being water soluble, THF might do the trick too with good stirring. I don't know if HOAc will interfer with the reduction.

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Bandil

When bromomethylating benzene, it is quite easy to purify it, as they simply add lots of water and extract with something non-polar. This is not possible in the case of 4-bromomethyl-2,5-dimethoxyamphetamine, as it will be in its acidic form after the reaction. As basification was not an option, it seems rather hard to isolate it.

Any bright ideas on how to solve this problem, without turning to chloromethylation?

Also, check out the early volumes of Organic Reactions. I remember one containing a chapter on chloromethylations, and several references to the chloromethylation of benzaldehydes are given there. I do know for sure there is a reference given for the chloromethylation of anisaldehyde with a 90% yield.

Bandil

I dont have the reference here, but i read in March - Advanced organic chemistry under the section "dehalogenation", that primary and secondary alkylhalogens are easily reduced/dehalogenated using Zinc in basically any acidic medium. Anyone have the reference handy? I dont have the book here, so i cannot post it before tomorrow...

lugh>Any trick to reading that article of yours? I just get a "picproxy.pl" file for download or something like that...

RegardsBandil

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Nicodem

I noticed that in this thread there has been a misunderstanding of an essential part of the DOMOM-DOMCl paper. The chloromethylation works only because the conditions and particularly the solvent (chlorobenzene) were such as to avoid the Pictet-Spengler reaction:"In order to synthesize a dimethoxyamphetamine with a chloromethyl group in position 4, it is necessary to avoid reactions with the highly reactive amino function."

The authors rationalize their choice, but they do it in such a way that it was impossible for me to understand it completely. So please, if there is somebee out there better skilled in English language he could make me slightly happier by decoding this:"The problems mentioned can be avoided using the Blanc reaction, if it is possible to find a dissolver for 2,5-dimethoxyamphetamine [17], which forms a hydrochloride salt under these working conditions, and is therefore insoluble in the known solvents for this reaction. A solventwhich is polar enough to dissolve the educt as hydrochloride is unsuitable in this Blanc reaction."[17] A.Shulgin in Pihkal (Ed..Joy),TransformPress, Barkley 1991, p. 53. Could someone with the printed version of PIHKAL please check what did they reffer to.

Maybe chlorobenzene can be easily substituted with toluene even though toluene can be also chloromethylated. But I think this side reaction would not be of great extent because the temperature "should not exceed 20°C" and the preparation done in 30 minutes. Furtermore the 2,5-Dimethoxyamphetamine is much more reactive than toluene. But I'm highly skeptic about using acetic acid as a solvent as proposed (even if it is for the bromomethylation). I think this would favor the Pictet-Spengler reaction yielding mostly the dimethoxy-methyl-tetrahidroisoquinoline (formic acid is the usual solvent for these Pictet-Spengler condensations!).

primary and secondary alkylhalogens are easily reduced/dehalogenated using Zinc in basically any acidic medium

When I proposed this Zn/AcOH system for the reduction of their "haloephedrines" in the Stimulants forum there was no response at all. It is funny how fanatically they stick with their red phosphorus/iodine .

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imp

General Reduction Procedure: The chloromethyl derivative (0.5g) in acetic acid (8 ml.) and water (2 ml.) was treated during 0.5 hr. with zinc dust (0.5 g.) and the reaction mixture heated on a steam bath for 2 hr. in all. The reaction mixture was then filtered and poured into cold water. The methyl derivative which separated was crystallized from dilute acetic acid. It was necessary to add hydrochloric acid (0.5 ml.) in the case of 7-methoxy-3,6-dichloromethyl-4-methylcoumarin to prevent the formation of the diacetoxymethyl derivative, and in the case of 7-hydroxy-2,8-dichloromethylcoumarin and 7-methoxy-3-chloromethyl-4-methylcoumarin-6-carboxylic acid to precipiatate the product.

-No yields given. But, SWIM has A LOT of other refs all stating that Zn/AcOH is a standard way of reducing benzylic chloromethyl compounds and offers good yields.

A solution of 1.05 g. of 9-methyl-10-iodomethyl-1,2-benzanthracene in 60 cc. of dioxane containing 3 cc. of hydrochloric acid was added a solution of 10 g. of stannous chloride crystals in 50 cc. of dioxane and 30 cc. of concentrated hydrochloric acid. It is necessary to use a very strongly acidic medium in order to avoid the formation of tar. After being refluxed gently for 5 minutes, the originally dark yellow solution became pale yellow; the solution was allowed to stand for one-half hour and then diluted with 500 cc. of water. After standing overnight to allow the precipitate hydrocarbon to coagulate, there was collected 0.71 g. (99%) of only faintly colored 9,10-dimethyl-1,2-benzanthracene.

JACS vol 80, pp1405 (1958)...

10-iodomethyl-9-anthranyl-propionic acid was dissolved in a boiling solution of 224g. of stannous chloride, 675ml concentrated hydrochloric acid and 1125 ml. of dioxane. The clear, light yellow solution was cooled and poured into 2250 ml. of water. The precipitated solid was collected and washed successively with 15% hydrochloric acid and water. Crystallization of the crude acid from ethyl acetate afforded 45.1g (85%) of the methyl derivative.

*Note that this same reduction technique is used to reduce 5-chloromethylvanillin in the synthesis of 5-Me-MDA on Rhodium's.

Reduction using Zinc in two-phase system of benzene and 2N NaOH---------------------------------------------------------------

SWIM will type it up if anybody is interested. Many chloromethyl compounds will be fine under basic conditions, but this obviously won't work for 2,5-DMA.HCl.

SWIM likes the idea of halomethylating 2,5-DMB. Reaction will occur at the activated 4-position, and the SnCl2/HCl reduction in dioxane would seem to be the most sure-fire method (the procedure having been used successfully on a benzaldehyde... see 5-Me-MDA).

To moo: SWIM read the Quelet paper too. Not sure why you say the main product is the diphenylalkane compound - diphenylalkanes only formed when a lack of HCl(aq & g) and not enough cooling were applied. Otherwise, Quelet was able to chloroethylate and propylate anisole in decent yields. Although he mentions chloropropylation is much more difficult to effect than chloroethylation. A simple method to DOET would be amazing.

Also iodomethylation is quite easy to effect as well. HI/paraformaldehyde will usually do the trick, or you can take the chloromethyl derivative, dissolve in some GAA and add HI... the iodomethyl derivative is much less soluble and will precipitate out. SWIM can provide literature if anyone is interested.

Nicodem

Imp, thanks for providing the examples of Zn/AcOH reductions. It was already time for somebody to do that.

SWIM likes the idea of halomethylating 2,5-DMB. Reaction will occur at the activated 4-position

The position 4 is deactivated by the -CHO group. The most probable position for an electrophilic attack on 2,5-diMeO-benzaldehide is the position 3 as it is activated by MeO- (ortho to it) and the less deactivated by -CHO (meta to it; the other meta position is ocupied by the 5-MeO).

moo

Imp: I was thinking along the lines of a H2SO4/KBr reaction with acetaldehyde and the aromatic. According to that article the diphenylmethane would be the product. But you are right about the actual reaction in that paper. The second article in that series suggests that using chloroethylation with dimethoxybenzenes might not be that easy, though, as they say that even the reaction products obtained from cresol methyl ethers are very labile and side reactions are unavoidable. On the other hand they also say that ortho-methoxy compounds are much more stabile than para-methoxy compounds. If it could be get to work somehow, it would be nice indeed!

One way to overcome the formation of diphenylmethanes would be the addition of the aromatic substrate to the reaction mixture (eg.

(Chimimanie: "Synthesis of 2-Chloromethyl-5-alkyl-DMB", Methods Discourse)), but depending on the substrate it might give rise to a disubstituted product, the reason why p-dimethoxybenzene is chloromethylated the other way around (

Nicodem: According to those rules 2,5-dimethoxybenzaldehyde would get brominated in the position 3 instead of the experimentally verified positions 4 and 6. It appears that sometimes those rules do not apply.